Integrating $\tan z$ Around a Circle of Radius 8

[Dustinsfl]

So if I wanted to integrate $\tan z$ around a circle of radius 8, I would need to identify all the singularities and then use
$$
\int_Cf=2\pi i\sum_{n = 1}^m a_i\text{Res}_{z_i}f,
$$
where $a_i = \eta(\gamma,z_i)$ the winding number.
Correct?

 

[chisigma]

Yes, that's correct... in case of a circle of course all the singularities have 'winding number' equal to 1...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

Yes, that's correct... in case of a circle of course all the singularities have 'winding number' equal to 1...

Kind regards

$\chi$ $\sigma$

So the only singularities would be at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ or do I have to consider and infinite amount of revolutions?

 

[chisigma]

In a circle of radius 8 centered in z=0 the singularities are $\displaystyle z= \pm \frac{\pi}{2}$, $\displaystyle z=\pm \frac{3 \pi}{2}$ and $\displaystyle z=\pm \frac{5 \pi}{2}$. Now You have to compute the residue of each singularity, the sum them and finally multiply by $\displaystyle 2 \pi\ i$...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

In a circle of radius 8 centered in z=0 the singularities are $\displaystyle z= \pm \frac{\pi}{2}$, $\displaystyle z=\pm \frac{3 \pi}{2}$ and $\displaystyle z=\pm \frac{5 \pi}{2}$. Now You have to compute the residue of each singularity, the sum them and finally multiply by $\displaystyle 2 \pi\ i$...

Kind regards

$\chi$ $\sigma$

So the answer is $-12\pi i$, correct?

 

[chisigma]

The function $\displaystyle \tan z$ around $\displaystyle z=(2n+1)\ \frac{\pi}{2}$ has the same Laurent expansion of the function $\displaystyle - \cot z$ around $\displaystyle z=0$ so that each residue is - 1 and the result is $\displaystyle - 12\ \pi\ i$...

Kind regards

$\chi$ $\sigma$